Momentum of a proton in de Broglie wavelength

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SUMMARY

The de Broglie wavelength of a proton is determined by its momentum, expressed as λ=h/p, where h is Planck's constant. When protons are accelerated from rest through a potential V, their momentum can be derived from the relationship p=mv, with kinetic energy given by mv²/2 = eV. The correct expression for the de Broglie wavelength of the proton is B) h/√(2meV).

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically de Broglie wavelength
  • Knowledge of momentum and kinetic energy equations
  • Familiarity with Planck's constant and its significance in quantum physics
  • Basic concepts of electric potential and energy conversion
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  • Study the derivation of de Broglie wavelength in quantum mechanics
  • Learn about the relationship between kinetic energy and potential energy in charged particles
  • Explore the implications of Planck's constant in quantum theory
  • Investigate the behavior of protons in electric fields and their acceleration
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Students of quantum mechanics, physics educators, and anyone interested in the behavior of subatomic particles under electric fields.

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Homework Statement



In quantum mechanics the de Broglie wavelength of an object depends
on its momentum according to λ=h/p where h is Planck's constant.
Protons of charge e and mass m are accelerated from rest through a
potential V. What is their de Broglie wavelength?

A) 2h/\sqrt{}meV
B) h/\sqrt{}2meV
C) h\sqrt{}meV
D) h/eV

Homework Equations



I don't know.

The Attempt at a Solution



I just don't know what I have to search for to solve it.
How can I express p with e and V?
 
Physics news on Phys.org
Hello there.
As you probably know p=mv. From conservation of energy u have mv^2/2 = eU, here you can solve for the velocity v. After that you can use it to end up with answer B)
 
Thanks.
 

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